Fidelity of quantum states

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In quantum information theory, fidelity is a measure of the "closeness" of two quantum states. It is not a metric on the space of density matrices.

Contents

[edit] Motivation

In probability theory, given two random variables p = (p1...pn) and q = (q1...qn) on the probability space X = {1,2...n}. The fidelity of p and q is defined to be the quantity

F(p,q) = \sum _i \sqrt{p_i q_i}.

In other words, the fidelity F(p,q) is the inner product of (\sqrt{p_1}, \cdots ,\sqrt{p_n}) and (\sqrt{q_1}, \cdots ,\sqrt{q_n}) viewed as vectors in Euclidean space. Notice that when p = q, F(p,q) = 1. In general, 0 \leq F(p,q) \leq 1.

Making the appropriate modification for the matricial notion of square root and mimicking the above definition give the fidelity of two quantum state.

[edit] Definition

Given two density matrices ρ and σ, the fidelity is defined by

F(\rho, \sigma) = \operatorname{Tr} (\sqrt{\sqrt{\rho} \sigma \sqrt{\rho}}).

By M½ of a positive semidefinite matrix M, we mean its unique positive square root given by the spectral theorem. The Euclidean inner product from the classical definition is replaced by the Hilbert-Schmidt inner product. When the states are classical, i.e. when ρ and σ commute, the definition coincides with that for probability distributions.

Notice by definition F is non-negative, and F(ρ,ρ) = 1. In the following section it will be shown that it can be no larger than 1.

[edit] Simple examples

[edit] Pure states

Consider pure states \rho = | \phi \rangle \langle \phi | and \sigma = | \psi \rangle \langle \psi |. Their fidelity is

F(\rho, \sigma) = \operatorname{Tr} [\rho \; \sigma \rho]^{\frac{1}{2}} = \operatorname{Tr} \; [ | \phi \rangle \langle \phi | \psi \rangle \langle \psi |\phi \rangle \langle \phi | ]^{\frac{1}{2}} = | \langle \phi | \psi \rangle |.

This is sometimes called the overlap between two states. If, say, |\phi\rangle is an eigenstate of an observable, and the system is prepared in | \psi \rangle, then F(ρ, σ)2 is the probability of the system being in state |\phi\rangle after the measurement.

[edit] Commuting states

Let ρ and σ be two density matrices that commute. Therefore they can be simultaneously diagonalized by unitary matrices, and we can write

\rho = \sum_i p_i | i \rangle \langle i | and \sigma = \sum_i q_i | i \rangle \langle i |

for some orthonormal basis \{ | i \rangle \}. Direct calculation shows the fidelity is

F(\rho, \sigma) = \sum_i \sqrt{p_i q_i}.

This shows that, heuristically, fidelity of quantum states is a genuine extension of the notion from probability theory.

[edit] Some properties

[edit] Unitary invariance

Direct calculation shows that the fidelity is preserved by unitary evolution, i.e.

\; F(\rho, \sigma) = F(U \rho \; U^*, U \sigma U^*)

for any unitary operator U.

[edit] Uhlmann's theorem

We saw that for two pure states, their fidelity coincides with the overlap. Uhlmann's theorem generalizes this statement to mixed states, in terms of their purifications:

Theorem Let ρ and σ be density matrices acting on Cn. Let ρ½ be the unique positive square root of ρ and

| \psi _{\rho} \rangle = \sum_{i=1}^n \rho^{\frac{1}{2}} | e_i \rangle \otimes | e_i \rangle \in \mathbb{C}^n \otimes \mathbb{C}^n

be a purfication of ρ (therefore {|ei >} is an orthonormal basis), then the following equality holds:

F(\rho, \sigma) = \max_{|\psi_{\sigma} \rangle} | \langle \psi _{\rho}| \psi _{\sigma} \rangle |

where | \psi _{\sigma} \rangle is a purification of σ. Therefore, in general, the fidelity is the maximum overlap between purifications.


Proof: A simple proof can be sketched as follows. Let |Ω > denote the vector

| \Omega \rangle= \sum | e_i \rangle \otimes | e_i \rangle

and σ½ be the unique positive square root of σ. We see that, due to the unitary freedom in square root factorizations and choosing orthonormal bases, an arbitrary purification of σ is of the form

| \psi_{\sigma} \rangle = ( \sigma^{\frac{1}{2}} V_1 \otimes V_2 ) | \Omega \rangle

where Vi's are unitary operators. Now we directly calculate

| \langle \psi _{\rho}| \psi _{\sigma} \rangle | = \langle \Omega | ( \rho^{\frac{1}{2}} \otimes I) ( \sigma^{\frac{1}{2}} V_1 \otimes V_2 ) | \Omega \rangle = \operatorname{Tr} ( \rho^{\frac{1}{2}} \sigma^{\frac{1}{2}} V_1 V_2 ).

But in general, for any square matrix A and unitary U, it is true that |Tr(AU)| ≤ Tr (A*A)½. Furthermore, equality is achieved if U* is the unitary operator in the polar decomposition of A. From this follows directly Uhlmann's theorem.

[edit] Consequences

Some immediate consequences of Uhlmann's theorem are

  • Fidelity is symmetric in its arguments, i.e. F (ρ,σ) = F (σ,ρ). Notice this is not obvious from the definition.
  • F (ρ,σ) lies in [0,1], by the Cauchy-Schwarz inequality.
  • F (ρ,σ) = 1 if and only if ρ = σ, since Ψρ = Ψσ implies ρ = σ.

[edit] References

  • A. Uhlmann The "Transition Probability" in the State Space of a *-Algebra. Rep. Math. Phys. 9 (1976) 273 - 279. PDF
  • R. Jozsa, Fidelity for mixed quantum states, Journal of Modern Optics, 1994, vol. 41, 2315-2323.
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